當點擊按鈕時，出現「無效」信息

-1

當我嘗試登錄時，我一直收到「無效的電子郵件或密碼」消息。我的錯在哪裏？你有解決這個問題的建議嗎？或者有沒有獲得有關此信息的來源？當點擊按鈕時，出現「無效」信息

這是mainactivity.java，這部分是關於登錄按鈕。當我點擊按鈕時，它不檢查數據庫獲取信息，也不會切換到另一個頁面;

private View.OnClickListener btnGirisListener = new View.OnClickListener() { 
    @Override 
    public void onClick(View v) { 
     email = etmail.getText().toString(); 
     password = etpassword.getText().toString(); 
     login(email, password); 
    } 
}; 
    private void login(final String email, String password){ 
     class LoginAsync extends AsyncTask<String,Void,String>{ 
      private Dialog loadingDialog; 

      @Override 
      protected void onPreExecute() { 
       super.onPreExecute(); 
       loadingDialog=ProgressDialog.show(MainActivity.this,"Please wait", "Loading.."); 
      } 

      @Override 
      protected String doInBackground(String... params) { 
       String mail = params[0]; 
       String pass = params[1]; 

       InputStream is = null; 
       List<NameValuePair>nameValuePairs = new ArrayList<NameValuePair>(); 
       nameValuePairs.add(new BasicNameValuePair("E-mail",mail)); 
       nameValuePairs.add(new BasicNameValuePair("Password", pass)); 
       String result = null; 

       try{ 
        HttpClient httpClient = new DefaultHttpClient(); 
        HttpPost httpPost = new HttpPost("http://******"); 
        httpPost.setEntity(new UrlEncodedFormEntity(nameValuePairs)); 

        HttpResponse response = httpClient.execute(httpPost); 

        HttpEntity entity = response.getEntity(); 

        is = entity.getContent(); 

        BufferedReader reader = new BufferedReader(new InputStreamReader(is, "UTF-8"), 8); 
        StringBuilder sb = new StringBuilder(); 

        String line = null; 
        while ((line = reader.readLine()) != null) 
        { 
         sb.append(line + "\n"); 
        } 
        result = sb.toString(); 
       } catch (ClientProtocolException e) { 
        e.printStackTrace(); 
       } catch (UnsupportedEncodingException e) { 
        e.printStackTrace(); 
       } catch (IOException e) { 
        e.printStackTrace(); 
       } 
       return result; 
      } 

      @Override 
      protected void onPostExecute(String result) { 
       String s = result.trim(); 
       loadingDialog.dismiss(); 
       if(s.equalsIgnoreCase("success")){ 
        Intent intent = new Intent(MainActivity.this, Giris.class); 
        intent.putExtra(E_MAIL, email); 
        finish(); 
        startActivity(intent); 
       }else{ 
        Toast.makeText(getApplicationContext(),"Invalid E-mail or Password.",Toast.LENGTH_SHORT).show(); 
       } 
      } 
     } 
     LoginAsync la = new LoginAsync(); 
     la.execute(email,password); 
    } 

};

我試圖更改php文件，但無法獲得任何解決方案。和uyegiris.php文件;

<?php 
define('DB_USER', "*******"); //Blanks are full. 
define('DB_PASSWORD', "*****"); 
define('DB_DATABASE', "*****"); 
define('DB_SERVER', "*****"); 

$con = mysqli_connect(DB_USER,DB_PASSWORD,DB_DATABASE,DB_PASSWORD); 

$Email = $_POST['Email']; 
$Sifre = $_POST['Sifre']; 

$sql = "select * from uyeler where Email='$Email' and Sifre='$Sifre'"; 

$res = mysqli_query($con,$sql); 
$check = mysqli_fetch_array($res); 

if(isset($check)){ 
    echo 'success'; 
}else{ 
    echo 'failure'; 
} 
mysqli_close($con); 

?>

來源

2017-08-24 Can Dirik

我認爲在PARAMS一些錯字錯誤，如設置在PHP端的移動結束「電子郵件」和「電子郵件」在移動端也使用'Sifre'而不是'Password'。 –

相同的密碼在PHP它是'Sifre'和在Android它''密碼' –

爲什麼你定義整個AsyncTask類在你的登錄方法。在登錄方法和登錄方法外定義AsyncTask只需調用 LoginAsync la = new LoginAsync（）; la.execute（email，password）; – Danger

這好像你已經做了在參數鍵一些錯字錯誤移動（發送）和PHP（接收），這是修正如下：

移動（發送）：

nameValuePairs.add(new BasicNameValuePair("Email",mail)); 
nameValuePairs.add(new BasicNameValuePair("Password", pass));

PHP（接收）：

$Email = $_POST['Email']; 
$Sifre = $_POST['Password'];

來源

2017-08-24 10:40:56

我已經完成了你所說的。但它仍然不起作用。我仍然收到「無效」消息。 –

@ CanDirik，嘗試打印帶有值的選擇查詢並在數據庫上運行該查詢並檢查它是否工作。 –

看起來像是在工作。 –

當點擊按鈕時，出現「無效」信息

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